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Rotating torque. A rectangular plate of mass \(M\), with sides \(a\) and \(b\), is rotated with angular velocity \(\omega\) about a fixed axis along a diagonal. Evaluate the rotating torque vector that the bearings must apply to the plate to hold it in this mode of rotation. Draw a good diagram showing the angular momentum vector. Express it as a vector.

Short Answer

Expert verified
Torque vector depends on the moment of inertia \(I_d = \frac{1}{12}M(a^2 + b^2)\) and angular velocity \(\omega\).

Step by step solution

01

Understand the System

We need to analyze a rectangular plate that is rotating about a fixed axis along its diagonal. The plate has mass \(M\) and dimensions \(a\) and \(b\), rotating with an angular velocity \(\omega\). The aim is to find the torque vector needed to maintain this rotation.
02

Determine the Moment of Inertia

The moment of inertia \(I\) for a rectangular plate rotating about an axis through its diagonal is computed by using the parallel axis theorem. The moment of inertia along one side is \(I_a = \frac{1}{12}M(b^2)\) and along the other side \(I_b = \frac{1}{12}M(a^2)\). For rotation about the diagonal, the moment of inertia is \(I_d = \frac{1}{2}(I_a + I_b) = \frac{1}{12}M(a^2 + b^2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque
Torque is like the rotational version of force. Just as force causes an object to move in a straight line, torque causes an object to rotate. It is calculated by multiplying the force applied with the distance from the point of rotation. In mathematical terms, the torque \( \tau \) is given by the product of the lever arm \( r \) and the force \( F \) applied, or \( \tau = r \times F \).

In our exercise of a rotating rectangular plate, torque is crucial because it helps maintain the rotation of the plate around its diagonal. Imagine twisting a jar lid. That's torque in action! Similarly, to keep the plate spinning, the bearings apply a torque that counteracts any forces trying to stop the rotation.
  • Key Point: Torque direction follows the right-hand rule. Curl your fingers in the direction of rotation, and your thumb points in the torque's direction.
  • Practical Insight: Without enough torque, the plate would slow down and eventually stop rotating.
Moment of Inertia
The moment of inertia is a measure of how difficult it is to change the rotational motion of an object. Think of it as the rotational equivalent of mass. Bigger or more spread out mass means more resistance to changes in rotational speed.

For our rectangular plate, the moment of inertia, \( I \), determines how much torque is needed to rotate it. The dimensions of the plate (sides \( a \) and \( b \)) and its mass \( M \) are crucial to its calculation. We use the formula \( I = \frac{1}{12} M (a^2 + b^2) \) for rotation about its diagonal. This tells us how the mass is distributed relative to the axis of rotation.
  • Understanding: A larger value of \( I \) means you need more torque to achieve the same angular acceleration as an object with a smaller \( I \).
  • Note: The moment of inertia depends not just on mass, but also on the shape and axis of rotation.
Angular Momentum
Angular momentum is like momentum, but for rotating objects. It describes how much motion a rotating object has. It's the product of moment of inertia \( I \) and angular velocity \( \omega \), expressed as \( L = I \omega \).

In our exercise, the plate has angular momentum as it spins around its diagonal with velocity \( \omega \). Angular momentum is a conserved quantity, meaning that in an isolated system, it doesn’t change over time. This is why once the plate starts spinning, it keeps going, unless something like friction applies a force.
  • Visualizing: To picture angular momentum, think of a spinning ice skater pulling in their arms to spin faster. That's them changing their moment of inertia to adjust their angular momentum.
  • Important: The direction of the angular momentum vector is perpendicular to the plane of rotation, dictated by the right-hand rule.

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Most popular questions from this chapter

Moment of inertia of solid sphere. Show that the moment of inertia about a diameter of a solid sphere is \(\frac{2}{5} M r^{2} .\) This can be simply done by considering the sphere to be a stack of circular disks of infinitesimal thickness fitting within a spherical bounding surface.

Parallel axis theorem. Beginning with the fact that the moment of inertia of a thin disk about a diametral axis is \(\frac{1}{4} m a^{2}\), employ the parallel axis theorem to prove that for a solid circular cylinder of mass \(M\), radius \(a\), and length \(L\), the moment of inertia about a transverse axis through the center of mass is \(M a^{2} / 4+M L^{2} / 12\)

Frictional torque. A heavy flywheel in the form of a solid cylinder of radius \(50 \mathrm{~cm}\), thickness \(20 \mathrm{~cm}\), and mass \(1200 \mathrm{~kg}\) rotates freely on bearings at an initial rate of \(150 \mathrm{rps}\). It is to be brought to rest by a friction brake, in which a brake shoe is pressed against the periphery of the flywheel with a force equivalent to a \(40-\mathrm{kg}\) weight. The coefficient of friction between the braking surfaces is \(0.4\) and is assumed to be independent of relative surface speed. (a) Through what angle will the flywheel turn in coming to rest if the brake is steadily applied? (b) How long will it be in coming to rest?

Gyroscope. A certain gyroscope consists of a solid cylinder with radius \(a=4 \mathrm{~cm} .\) It is supported by a massless stem whose tip is pivoted freely at a point \(5 \mathrm{~cm}\) from the center of mass of the cylinder. It is observed to be moving in steady precession at an angle of inclination from vertical, and the precession occurs at one complete circular excursion every \(3 \mathrm{~s}\). Evaluate the angular velocity of spin of the gyroscope about its own axis.

Unbalanced rigid body. A thin rim or hoop of mass \(M\) and radius \(R\) is mounted with massless spokes so as to rotate freely in the vertical plane about a horizontal axis through its center. A particle of mass \(m\) is fastened to the rim, causing the system to hang at rest with \(m\) at the bottom. Find the frequency of small oscillations. Also find the maximum angular velocity attained if the system is released from a stationary condition with \(m\) at the top.

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